Question

How do you find the exact value of `(sin2theta)/(1+cos2theta)=4` in the interval `0<=theta<2pi`?

Answer 1

`theta~=0,42pi or theta~=1,42pi`

Explanation:

Since

`sin2theta=2sinthetacostheta`

and

`cos2theta=cos^2theta-sin^2theta`

you can substitute in the give equation and get:

`(2sinthetacostheta)/(1+cos^2theta-sin^2theta)=4`

Then, since `1-sin^2theta=cos^2theta`,

you can substitute and get:

`(2sinthetacostheta)/(cos^2theta+cos^2theta)=4`

`(cancel2sinthetacancelcostheta)/(cancel2cos^cancel2theta)=4`

`tantheta=4`

`theta=arctan4`

`theta~=0,42pi or theta~=1,42pi`

Answer 2

two solutions
`theta_1=tan^(-1)(1/4)`, `theta_2=tan^(-1)(1/4)+pi`,

Explanation:

`sin2theta=2sinthetacostheta`
`cos2theta=cos^2theta-sin^2theta`

By replacing them into the equation we get
`2sinthetacostheta=4+4cos^2theta-4sin^2theta`

besides in order to make the equation homogenous we can replace 4 with `4sin^2theta+4cos^2theta` so that the equation becomes

`2sinthetacostheta=4cos^2theta+4sin^2theta+4cos^2theta-4sin^2theta`
ruling out the opposite terms we get

`8cos^2theta-2sinthetacostheta=0`

and by collecting the common factor

`2costheta(4costheta -sintheta)=0`

It is satisfied when `costheta=0` (for `theta=pi/2+kpi`)

and when `sintheta=4costheta`
that by dividing both sides for `costheta` (provided it is different from 0). It turns into
`tantheta=1/4`
that is satisfied for `theta=tan^(-1)(1/4)+kpi`

the other two possible solutions found for the first factor (`theta=pi/2`, `theta=3/2pi`) can not be taken as they break the existence condition for the given fractional equation